l Advanced Subsidiary Paper 1: Pure athematics PAPER B ark Scheme 1 Any reasonable explanation. For example, the student did not correctly find all values of x which satisfy cosx. Student should have subtracted 150 from 60 first, and then divided by. N.B. If insufficient detail is given but location of error is correct then mark can be awarded from working in part (b). x = 75 x = 105 (1 mark) ( marks) marks NOTE: 1a: Award the mark for a different explanation that is mathematically correct, provided that the explanation is clear and not ambiguous. akes an attempt to use Pythagoras theorem to find a. For example, 4 7 seen. 65 Displays the correct final answer. 4i 7j 1 65 ( marks)
Attempt to multiply the numerator and denominator by k(8 ). For example, 6 4 8 8 8 Attempt to multiply out the numerator (at least terms correct). 48 18 4 Attempt to multiply out the denominator (for example, terms correct but must be rational or 64 seen or implied). 64 8 8 p and q stated or implied (condone if all over 61). 44 14 or 61 61 44 14 p, q 61 61
4a akes an attempt to expand the binomial expression 1 x (must be terms in x 0, x 1, x, x and at least correct). x x x x x 1 1 0 < x x > 0* as required. * 4b Picks a number less than or equal to zero, e.g. x = 1, and attempts a substitution into both sides. For example, 1 1 1 1 1 1 1 Correctly deduces for their choice of x that the inequaltity does not hold. For example, 0 ( marks) 6 marks
5 Uses laws of indices correcty at least once anywhere in solution (e.g. 1 x x 1 or 1 x x or akes an attempt at integrating x x x seen or implied). 1 h'( x) 15x 40x Raising at least one x power by 1 would constitute an attempt. Fully correct integration. 5 1 6x 80x (no need for +C here). akes an attenpt to substitute (4, 19) into the integrated expression. For example, 5 1 19 6 4 80 4 C is seen. Finds the correct value of C. C = 1 States fully correct final answer 5 h( x) 6x 80 x 1 or any equivalent form. (6 marks) NOTES: Award all 6 marks for a fully correct final answer, even if some working is missing.
6 States sin xcos x 1 or implies this by making a substitution. 8 7cos x 6 1 cos x Simplifies the equation to form a quadratic in cos x. 6cos x 7cos x 0 Correctly factorises this equation. x x cos cos 1 0 or uses equivalent method for solving quadratic (can be implied by correct solutions). Correct solution. cos x or 1 Finds one correct solution for x. (48.,60, 11.8 or 00 ). Finds all other solutions to the equation. (6 marks)
7a States or implies the expansion of a binomial expression to the 8th power, up to and including the x term. 8 8 8 8 7 8 6 8 5 0 1 ( a b) C a C a b C a b C a b... or 8 8 7 6 5 ( a b) a 8a b 8a b 56 a b... Correctly substitutes 1 and x into the formula: 8 8 7 6 5 (1 x) 1 81 x 81 x 561 x... akes an attempt to simplify the expression ( correct coefficients (other than 1) or both 9x and 7x ). 8 8 (1 x) 1 4x 89x 56 7 x... dep States a fully correct answer: 8 (1 x) 1 4x 5x 151 x... 7b States x = 0.01 or implies this by attempting the substitution: 14 0.01 5 0.01 151 0.01... Attempts to simplify this expression ( calculated terms correct): 1 + 0.4 + 0.05 + 0.00151 1.6671 = 1.667 (5 s.f.) ( marks) 7 marks
8a Attempt to find intersection with x-axis. For log xa 0 example, Solving x a 9 log9 0 to find x = a + 1, so coordinates of x-intercept are ( a + 1, 0) oe Substituting x = 0 to derive y log 9 x a coordinates of y-intercept are 0,log 9 x a, so Asymptote shown at x = a stated or shown on graph. Increasing log graph shown with asymptotic behaviour and single x-intercept. Fully correct graph with correct asymptote, all points labelled and correct shape. (6 marks) 8b log x a log x a seen. 9 9 The graph of y log 9 x a is a stretch, parallel to the y-axis, scale factor, of the graph of y log 9 x a. ( marks) 8 marks NOTES: 8a: Award all 5 points for a fully correct graph with asymptote and all points labelled, even if all working is not present
9a Graph of y = x + 5 drawn. Graph of y + x = 6 drawn. Graph of y = drawn onto the coordinate grid and the triangle correctly shaded. 9b Attempt to solve y = x + 5 and y + x = 6 simultaneously for y. y =.4 Base of triangle =.5 ( marks) Area of triangle = 1 (.4 ).5 Area of triangle is.45 (units ). (5 marks) 8 marks NOTES: 9b: It is possible to find the area of triangle by realising that the two diagonal lines are perpendicular and therefore finding the length of each line using Pythagoras theorem. Award full marks for a correct final answer using this method. In this case award the second and third accuracy marks for finding the lengths.45 and 9.8
10a a 1 States that tan θ or θ tan (if θ shown on diagram sign must be consistent with this). Finds.7 (must be negative). ( marks) 10b a 10c a akes an attempt to use the formula F = ma Finds p = 10 Note: 8 p 6 p 10 Finds q = Note: 10 q 6 q ( marks) Attempt to find R (either 6(i j) or 8i 10 j '10' i ' ' j). akes an attempt to find the magnitude of their resultant force. For example, R '18' '1' 468 Presents a fully simplified exact final answer. R 6 1 ( marks) 8 marks
11 Attempts to differentiate. f ( x) x 14x 4 States or implies that the graph of the gradient function will cut the x-axis when f ʹ(x) = 0 x x x f ( ) 0 14 4 0 Factorises f ʹ(x) to obtain (x 4)( x 6) 0 x = 4, x = 6 States or implies that the graph of the gradient function will cut the y-axis at f ʹ(0). Substitutes x = 0 into f ʹ(x) Gradient function will cut the y-axis at (0, 4). Attempts to find the turning point of f ʹ(x) by differentiating (i.e. finding f ʹʹ(x)) 7 f ( x) 0 6x 14 0 x Substitutes 7 x into f ʹ(x) to obtain 11 y A parabola with correct orientation with required points correctly labelled. ft ft (9 marks) NOTES: A mistake in the earlier part of the question should not count against the students for the last part. If a student sketches a parabola with the correct orientation correctly labelled for their values, award the final mark. Note that a fully correct sketch without all the working but with all points clearly labelled implies 8 marks in this question.
1a a Equates the curve and the line. x 8x 0 x 6 Simplifies and factorises. (x 7)(x ) = 0 (or uses other valid method for solving a quadratic equation). Finds the correct coordinates of A. A(, 8). Finds the correct coordinates of B. B(7, 1). 1b a akes an attempt to find the area of the trapezium bounded by x =, x = 7, the x-axis and the line. 5 8 1 or For example, Correct answer. Area = 5.5 o.e. 7 ( x 6)dx seen. ( marks) 1c a 7 x x x. ( 8 0)d akes an attempt to find the integral. Raising at least one x power by 1 would constitute an attempt. Correctly finds 1 4 0 x x x 7 4 8 akes an attempt to substitute limits into the definite integral. 196 140 16 40 Correct answer seen. 95 or 1.6 oe seen. (5 marks) 1d a Understands the need to subtract the two areas. ( 5.5 1.6 ) 0.8 units seen (must be positive). ( marks) 1 marks NOTES: 1a: If A0A0, award for full solution of quadratic equation (i.e. x =, x = 7).
1a a 1b a 1c a Student completes the square twice. Condone sign errors. x y 4 16 5 5 1 0 x y 4 5 40 So centre is (4, 5) and radius is 40 Substitutes x = 10 into equation (in either form). 10 8 10 y 10y 1 0 or y Rearranges to term quadratic in y 10 4 5 40 y 10y 1 0 (could be in completed square form y 5 4) Obtains solutions y =, y = 7 (must give both). Rejects y = 7 giving suitable reason (e.g. 7 < 5) or it would be below the centre or AQ must slope upwards o.e. m AQ ( 5) 1 = 10 4 m (i.e. 1 over their l m AQ ) Substitutes their Q into a correct equation of a line. For example, 10 b or y x 10 y = x + 7 ( marks) ft
1d a 1e a 6 AQ o.e. (could just be in coordinate form). AP o.e. so student concludes that point P has coordinates (, 1). 6 Substitutes their P and their gradient 1 ( m from 5c) into a correct equation of a line. For example, 1 1 1 1 y x PA 40 b 1 or y1 x Uses Pythagoras theorem to find AQ 40 EP. 9 Area of EPA = 1 40 40 (could be in two parts). 9 Area = 0 19 marks